# Chemistry

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Calculate the percent ionizationof a 0.15 M benzoic acid solution in pure water Nd also in a solution containing 0.10M sodium benzoate. Why does the percent ionization differ significantly in the two solutions?

• Chemistry -

Call benzoic acid HB and sodium benzoate NaB.
.............HB ==> H^+ + B^-
initial....0.15M....0.....0
change.......-x......x.....x
equil.....0.15-x......x....x
I used 6.14E-5 for Ka.
6.14E-5 = (H^+)(B^-)/(HB)
6.14E-5 = (x)(x)/(0.15-x)
x = (H^+) = about 0.003 which is (0.003/0.15)*100 = about 2% ionized.
You can clean that up if you wish.

When we add the NaB to the mixture, it changes only the B part of the Ka expression.
6.14E-4 = (x)(x+0.1)/(0.15-x)
Solve for x = 9.2E-5 = (H^+) and
%ion = (9E-5/0.15)*100 = about 0.06%. Again you can clean it up if you wish. Here is the scoop on why? Remember the common ion effect? In this case the benzoate ion, from the sodium benzoate or NaB, is the common ion. Look at what happens with Le Chatelier's Principle.
.......HB ==> H^+ + B^-
Now if we add B^-,as in NaB,it shifts the equilibrium to the left which obviously means less ionization. That's the whole idea behind buffered solutions. A weak acid, such as benzoic acid, and a salt of the weak acid, such as sodium benzoate, pushes the ionization far to the left.Now you can add a strong acid (which is consumed by the benzoate base) or a strong base (which is consumed by the benzoic acid) and the pH of the solution stays essentially the same.

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